Units and Measurements Previous Year Questions — JEE Main Physics
50 past-year Units and Measurements questions from JEE Main (Physics) — page 3.
Q9 (2025)
Match the LIST-I with LIST-II
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">LIST-I</th><th style="border: 1px solid #888; padding: 8px;">LIST-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Gravitational constant</td><td style="border: 1px solid #888; padding: 8px;">I. $[LT^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Gravitational potential energy</td><td style="border: 1px solid #888; padding: 8px;">II. $[L^2T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Gravitational potential</td><td style="border: 1px solid #888; padding: 8px;">III. $[ML^2T^{-2}]$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Acceleration due to gravity</td><td style="border: 1px solid #888; padding: 8px;">IV. $[M^{-1}L^3T^{-2}]$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- A-IV, B-III, C-II, D-I
- A-III, B-II, C-I, D-IV
- A-II, B-IV, C-III, D-I
- A-I, B-III, C-IV, D-II
### Related Formula
Newton's Law of Gravitation:
$$F = G\frac{m_1 m_2}{r^2} \implies G = \frac{Fr^2}{m_1 m_2}$$
Potential Energy:
$$U = mgh \quad [\text{Work}]$$
Potential:
$$V = \frac{W}{m}$$
Acceleration:
$$g = \frac{\text{Velocity}}{\text{Time}}$$
### Core Logic
Let's perform dimensional analysis for each item:
1. **A. Gravitational constant ($G$)**:
$$[G] = \frac{[F][r^2]}{[M^2]} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$$
Matches with **IV**.
2. **B. Gravitational potential energy ($U$)**:
$$[U] = \text{Dimensions of Work} = [ML^2T^{-2}]$$
Matches with **III**.
3. **C. Gravitational potential ($V$)**:
$$[V] = \frac{[\text{Energy}]}{[M]} = \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]$$
Matches with **II**.
4. **D. Acceleration due to gravity ($g$)**:
$$[g] = [\text{Acceleration}] = [LT^{-2}]$$
Matches with **I**.
### Step 1: Match and Selection
Let's align our matches:
- A $\rightarrow$ IV
- B $\rightarrow$ III
- C $\rightarrow$ II
- D $\rightarrow$ I
This sequence matches option (1).
### Pattern Recognition
To save precious exam time on match-the-column questions, start with the easiest dimensional terms first. You know Acceleration due to gravity is $g \rightarrow [LT^{-2}]$ (D-I) and energy is $[ML^2T^{-2}]$ (B-III). Looking at the options, only Option 1 matches this sequence immediately!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Class 11 Physics: Gravitation
Q17 (2025)
A person measures mass of 3 different particles as $435.42\mathrm{~g}$, $226.3\mathrm{~g}$ and $0.125\mathrm{~g}$. According to the rules for arithmetic operations with significant figures, the additions of the masses of 3 particles will be.
- $661.845\mathrm{~g}$
- $662\mathrm{~g}$
- $661.8\mathrm{~g}$
- $661.84\mathrm{~g}$
### Related Formula
Significant Figures Rule for Addition/Subtraction:
The final result must be rounded off to keep only as many decimal places as there are in the measurement with the **least** number of decimal places.
### Core Logic
Let's look at the decimal places of each measurement:
- $435.42\mathrm{~g}$ has **2 decimal places**.
- $226.3\mathrm{~g}$ has **1 decimal place**.
- $0.125\mathrm{~g}$ has **3 decimal places**.
The minimum number of decimal places is **1 decimal place** (from $226.3\mathrm{~g}$).
### Step 1: Addition and Rounding
First, perform the standard mathematical addition:
$$\text{Sum} = 435.42 + 226.3 + 0.125 = 661.845\mathrm{~g}$$
Now, round this raw sum off to **1 decimal place**:
- The digit after tenths place is 4 ($4 < 5$), so we round down.
- Net rounded sum = $661.8\mathrm{~g}$.
### Pattern Recognition
Bust the common myth: Addition depends on the least number of decimal places, whereas multiplication/division depends on the least number of significant figures. Always distinguish between these two rules during exams!
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements: Error Analysis and Significant Figures
Q4 (2025)
Given below are two statements:
Statement (I): The dimensions of Planck's constant and angular momentum are same.
Statement (II): In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.
In the light of the above statements, choose the most appropriate answer from the options given below:
- Both Statement I and Statement II are correct
- Statement I is incorrect but Statement II is correct
- Statement I is correct but Statement II is incorrect
- Both Statement I and Statement II are incorrect
### Related Formula
$$E = hf \implies [h] = \frac{[E]}{[f]} = \frac{\text{M}\text{L}^2\text{T}^{-2}}{\text{T}^{-1}} = \text{M}\text{L}^2\text{T}^{-1}$$
$$L = mvr \implies [L] = \text{M} \cdot (\text{L}\text{T}^{-1}) \cdot \text{L} = \text{M}\text{L}^2\text{T}^{-1}$$
$$L = \frac{nh}{2\pi}$$
### Core Logic
Statement I: Comparing the dimensional formula of Planck's constant ($h$) and angular momentum ($L$), both are identical $[\text{M}\text{L}^2\text{T}^{-1}]$. Hence, Statement I is correct.
Statement II: According to Bohr's second postulate, angular momentum is an integral multiple of $\frac{h}{2\pi}$, not an integral multiple of $h$. Hence, Statement II is incorrect.
### Pattern Recognition
Watch out for exact definitions in standard postulates. Bohr's model requires angular momentum to be quantized in units of $\hbar = \frac{h}{2\pi}$, making statement II a classic trap.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Class 12 Physics: Atoms
Q13 (2025)
For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm. The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm):
- 0.001
- 0.002
- 0.0005
- 0.0025
### Related Formula
$$1\text{ MSD} = \frac{\text{Total Length}}{\text{Total Divisions}}$$
$$\text{Least Count (LC)} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} \cdot \left(1 - \frac{24}{25}\right)$$
### Core Logic
Calculate the value of one main scale division:
$$1\text{ MSD} = \frac{15\text{ cm}}{300} = 0.05\text{ cm}$$
Given that $25\text{ VSD} = 24\text{ MSD}$, we find:
$$1\text{ VSD} = \frac{24}{25}\text{ MSD}$$
### Step 1: Compute Least Count
$$\text{LC} = 1\text{ MSD} \cdot \left(\frac{1}{25}\right) = \frac{0.05\text{ cm}}{25} = 0.002\text{ cm}$$
### Pattern Recognition
Least count calculation is conventionally $\text{LC} = \frac{1\text{ MSD}}{N}$ where $N$ represents the total count of divisions on the vernier scale whenever $(N-1)\text{ MSD} = N\text{ VSD}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Q20 (2025)
In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of $[M^{p}L^{Q}T^{R}A^{S}]$. The value of P and Q are :
- -1, 0
- -1, 1
- 1, -1
- 0, -1
### Related Formula
Electric Dipole Moment: $P_e = q \cdot d \implies [P_e] = \text{A}\cdot\text{T}\cdot\text{L}$
Magnetic Dipole Moment: $M_m = I \cdot A \implies [M_m] = \text{A}\cdot\text{L}^2$
### Core Logic
Take the dimensional ratio:
$$\left[\frac{P_e}{M_m}\right] = \frac{\text{L}\text{T}\text{A}}{\text{L}^2\text{A}} = \text{L}^{-1}\text{T} = \text{M}^0\text{L}^{-1}\text{T}^1\text{A}^0$$
Comparing powers with $[\text{M}^P\text{L}^Q\text{T}^R\text{A}^S]$, we find:
$P = 0$
$Q = -1$
### Pattern Recognition
Dipole units match up with fundamental currents and charge definitions. Notice that current $A$ cancels out completely, leaving a simple geometric spatial ratio.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Units and Measurements
Class 12 Physics: Moving Charges and Magnetism
